Arrange the following substances in the incre | vedclass

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Question

(D) The oxidation numbers of $Mn$ in the given compounds are calculated as follows:$MnCl_2$: $x + 2(-1) = 0 \Rightarrow x = +2$$MnO_2$: $x + 2(-2) = 0

Vedclass · Accepted Answer

$MnCl_2 < Mn(OH)_3 < MnO_2 < KMnO_4$

Vedclass · Answer

(D) The oxidation numbers of $Mn$ in the given compounds are calculated as follows:$MnCl_2$: $x + 2(-1) = 0 \Rightarrow x = +2$$MnO_2$: $x + 2(-2) = 0 \Rightarrow x = +4$$Mn(OH)_3$: $x + 3(-1) = 0 \Rightarrow x = +3$$KMnO_4$: $1(+1) + x + 4(-2) = 0 \Rightarrow x = +7$Thus,the increasing order of the oxidation number of $Mn$ is: $MnCl_2 (+2) < Mn(OH)_3 (+3) < MnO_2 (+4) < KMnO_4 (+7)$.

Arrange the following substances in the increasing order of the oxidation number of $Mn$: $MnCl_2, MnO_2, Mn(OH)_3, KMnO_4$.

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